Question: A regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let $R$ be the region outside the hexagon, and let $S = \left\lbrace\frac{1}{z} \ | \ z \in R\right\rbrace$.  Find the area of $S.$
We can compute that the side length of the hexagon is $\frac{1}{\sqrt{3}}.$  Then one side of the hexagon is parameterized by
\[\frac{1}{2} + ti,\]where $-\frac{1}{2 \sqrt{3}} \le t \le \frac{1}{2 \sqrt{3}}.$

[asy]
unitsize (4 cm);

pair A, B, C, D, E, F;

A = 1/sqrt(3)*dir(30);
B = 1/sqrt(3)*dir(30 - 60);
C = 1/sqrt(3)*dir(30 - 2*60);
D = 1/sqrt(3)*dir(30 - 3*60);
E = 1/sqrt(3)*dir(30 - 4*60);
F = 1/sqrt(3)*dir(30 - 5*60);

draw(A--B--C--D--E--F--cycle);
draw((-0.7,0)--(0.7,0));
draw((0,-0.7)--(0,0.7));

dot("$\frac{1}{2} + \frac{i}{2 \sqrt{3}}$", (1/2,1/(2*sqrt(3))), dir(0));
dot("$\frac{1}{2} - \frac{i}{2 \sqrt{3}}$", (1/2,-1/(2*sqrt(3))), dir(0));
[/asy]

Let $a + bi$ be a point on this side.  Then
\[x + yi = \frac{1}{a + bi} = \frac{a - bi}{a^2 + b^2} = \frac{\frac{1}{2} - ti}{\frac{1}{4} + t^2},\]so $x = \frac{\frac{1}{2}}{\frac{1}{4} + t^2}$ and $y = -\frac{t}{\frac{1}{4} + t^2}.$

We eliminate $t,$ to see what this point traces as $t$ varies.  Dividing these equations, we get
\[\frac{y}{x} = -2t,\]so $t = -\frac{y}{2x}.$  Substituting into the first equation, we get
\[x = \frac{\frac{1}{2}}{\frac{1}{4} + \frac{y^2}{4x^2}}.\]This simplifies to $x^2 + y^2 = 2x.$  Completing the square in $x,$ we get
\[(x - 1)^2 + y^2 = 1.\]This represents the circle centered at 1 with radius 1.

Hence, as $t$ varies over $-\frac{1}{2 \sqrt{3}} \le t \le \frac{1}{2 \sqrt{3}},$ $x + yi$ traces an arc of this circle.  Its endpoints are $\frac{3}{2} + \frac{\sqrt{3}}{2} i$ and $\frac{3}{2} - \frac{\sqrt{3}}{2} i.$  We can check that this arc is $120^\circ.$

[asy]
unitsize (4 cm);

pair A, B, C, D, E, F, P, Q;
path foo;
real t;

A = 1/sqrt(3)*dir(30);
B = 1/sqrt(3)*dir(30 - 60);
C = 1/sqrt(3)*dir(30 - 2*60);
D = 1/sqrt(3)*dir(30 - 3*60);
E = 1/sqrt(3)*dir(30 - 4*60);
F = 1/sqrt(3)*dir(30 - 5*60);

t = 1/(2*sqrt(3));
foo = (1/2/(1/4 + t^2),-t/(1/4 + t^2));
Q = (1/2/(1/4 + t^2),-t/(1/4 + t^2));

t = -1/(2*sqrt(3));
foo = (1/2/(1/4 + t^2),-t/(1/4 + t^2));
P = (1/2/(1/4 + t^2),-t/(1/4 + t^2));

for (t = -1/(2*sqrt(3)); t <= 1/(2*sqrt(3)); t = t + 0.01) {
  foo = foo--(1/2/(1/4 + t^2),-t/(1/4 + t^2));
}

draw(foo,red);
draw(A--B--C--D--E--F--cycle);
draw((-1,0)--(2.5,0));
draw((0,-1)--(0,1));
draw((1,0)--P,dashed);
draw((1,0)--Q,dashed);

label("$\frac{3}{2} - \frac{\sqrt{3}}{2} i$", Q, S);
label("$\frac{3}{2} + \frac{\sqrt{3}}{2} i$", P, N);

dot("$\frac{1}{2} + \frac{i}{2 \sqrt{3}}$", (1/2,1/(2*sqrt(3))), dir(0));
dot("$\frac{1}{2} - \frac{i}{2 \sqrt{3}}$", (1/2,-1/(2*sqrt(3))), dir(0));
dot(P,red);
dot(Q,red);
dot("$1$", (1,0), SW);
[/asy]

By symmetry, the rest of the boundary of $S$ can be obtain by rotating this arc by multiples of $60^\circ.$

[asy]
unitsize(2 cm);

path foo = arc((1,0),1,-60,60);
int i;

for (i = 0; i <= 5; ++i) {
  draw(rotate(60*i)*(foo),red);
	draw(rotate(60*i)*(((1,0) + dir(-60))--(1,0)--((1,0) + dir(60))));
	dot(rotate(60*i)*((1,0)));
  draw(rotate(60*i)*((0,0)--(1,0)--dir(60)));
}

for (i = 0; i <= 5; ++i) {
	dot(rotate(60*i)*((1,0) + dir(60)),red);
}
[/asy]

We can divide $S$ into 12 equilateral triangles with side length 1, and six $120^\circ$-sectors with radius 1, so the area of $S$ is
\[12 \cdot \frac{\sqrt{3}}{4} + 6 \cdot \frac{1}{3} \cdot \pi = \boxed{3 \sqrt{3} + 2 \pi}.\]Here are some alternative ways to derive the arc of the circle:

Alternative 1: Let $w = \frac{1}{z},$ where the real part of $z$ is $\frac{1}{2}.$  Write $w = r \operatorname{cis} \theta.$  Then
\[\frac{1}{z} = \frac{1}{w} = \frac{1}{r \operatorname{cis} \theta} = \frac{1}{r} \operatorname{cis} (-\theta) = \frac{\cos \theta - i \sin \theta}{r},\]so $\frac{\cos \theta}{r} = \frac{1}{2},$ or $r = 2 \cos \theta.$

If $x + yi = w = r \operatorname{cis} \theta = r \cos \theta + i \sin \theta,$ then
\[x^2 + y^2 = r^2 = 2r \cos \theta = 2x,\]so $(x - 1)^2 + y^2 = 1.$

Alternative 2: Let $w = \frac{1}{z},$ where the real part of $z$ is $\frac{1}{2}.$  Then $z$ is equidistant from 0 and 1 (the line $x = \frac{1}{2}$ is the perpendicular bisector of 0 and 1), so
\[|z| = |z - 1|.\]Dividing both sides by $z,$ we get
\[\left| 1 - \frac{1}{z} \right| = 1,\]so $|w - 1| = 1.$  Thus, $w$ lies on the circle centered at 1 with radius 1.